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Spring再接触 自动装配

时间:2018-09-09 14:52:32      阅读:123      评论:0      收藏:0      [点我收藏+]

标签:mod   1.0   pack   turn   xmla   pre   system   span   name   

UserDaoImpl

package com.bjsxt.dao.impl;

import com.bjsxt.dao.UserDAO;
import com.bjsxt.model.User;


public class UserDAOImpl implements UserDAO {
    
    private int daoId;
    
    public int getDaoId() {
        return daoId;
    }

    public void setDaoId(int daoId) {
        this.daoId = daoId;
    }

    public void save(User user) {
        //Hibernate
        //JDBC
        //XML
        //NetWork
        System.out.println("user saved!");
    }
    
    @Override
    public String toString() {
        return "daoId=" + daoId;
    }

}

beans.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
           http://www.springframework.org/schema/beans/spring-beans-2.5.xsd"
          >

  <bean name="userDAO" class="com.bjsxt.dao.impl.UserDAOImpl">
      <property name="daoId" value="1"></property>
  </bean>
  
  <bean name="userDAO2" class="com.bjsxt.dao.impl.UserDAOImpl">
      <property name="daoId" value="2"></property>
  </bean>
                              //default是指在表头指定default 然后这里设置为default就可以按照表头指定的来进行自动装配
  <bean id="userService" class="com.bjsxt.service.UserService" scope="prototype" autowire="byType">   //按照name的话就会自动匹配Userdao 按照类型只能有一个属性才行
  </bean>
  

</beans>

testadd

    @Test
    public void testAdd() throws Exception {
        ApplicationContext ctx = new ClassPathXmlApplicationContext("beans.xml");
        
        
        UserService service = (UserService)ctx.getBean("userService");
        
        System.out.println(service.getUserDAO());
        
        
        
    }

 

Spring再接触 自动装配

标签:mod   1.0   pack   turn   xmla   pre   system   span   name   

原文地址:https://www.cnblogs.com/frankzone/p/9613260.html

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